Math · Calculus I · Concept

The Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links derivatives and integrals. Part 1: if f is continuous and F(x) = ∫ₐˣ f(t) dt, then F′(x) = f(x). Part 2: if F is any antiderivative of a continuous f on [a, b], then the integral of f from a to b is F(b) − F(a).

Part 1: accumulated area has slope f

Fix a and let F(x) be the signed area under f from a to x. Moving x a little adds a thin strip of height about f(x), so the area grows at rate f(x). Where f is negative, the strips subtract and F decreases.

\frac{d}{d x} \int_{a}^{x} f (t) d t = f (x)

Why the slope equals the height

F(x + h) − F(x) is the integral over the short interval from x to x + h. Dividing by h gives the average height of f there, and at a point where f is continuous that average tends to f(x) as h → 0.

Part 2: evaluate with any antiderivative

To evaluate a definite integral, find any antiderivative F and subtract its values at the endpoints, upper minus lower. The constant C cancels in the subtraction, so it can be left out.

\int_{a}^{b} f (x) d x = F (b) - F (a)

Why the subtraction works

The accumulation function and any antiderivative F have the same derivative, so they differ by a constant. The accumulation function is 0 at a, so it equals F(x) − F(a), and at x = b that is F(b) − F(a).

Net change

Applied to a rate, Part 2 says the integral of a rate of change is the net change in the quantity. Integrating a velocity gives displacement; integrating a flow rate gives the net amount added.

\int_{a}^{b} F^{'} (t) d t = F (b) - F (a)

A variable upper limit and the chain rule

If the upper limit is a function g(x), Part 1 combines with the chain rule: evaluate f at g(x), then multiply by g′(x).

\frac{d}{d x} \int_{a}^{g (x)} f (t) d t = f (g (x)) g^{'} (x)

\begin{array}{c} \frac{d}{d x} \int_{a}^{g (x)} f (t) d t \\ = f (g (x)) g^{'} (x) \end{array}

Average value

Dividing an integral by the length of its interval gives the average value of the function: the height of the rectangle on [a, b] with the same area. The Mean Value Theorem for integrals says that if f is continuous, it reaches its average value at some point of the interval.

f_{avg} = \frac{1}{b - a} \int_{a}^{b} f (x) d x

Common mistakes

Confusing F(x), the accumulated area, with F′(x) = f(x), the current height.
Subtracting in the wrong order: the integral from a to b is F(b) − F(a), upper limit first.
Forgetting the chain-rule factor: the derivative of ∫₀^(x²) cos t dt is 2x cos(x²), not cos(x²).
Using Part 2 across a discontinuity: ∫ from −1 to 1 of 1/x² dx is not −2. The integrand is unbounded at 0 and the integral diverges.
Reading net change as total change: a displacement of zero can hide a lot of travel.

Key terms

Fundamental theorem of calculus: The link between derivatives and integrals: if F is an antiderivative of a continuous f, then ∫ₐᵇ f(x) dx = F(b) − F(a). It also says d/dx ∫ₐˣ f(t) dt = f(x).
Accumulation function: A function defined by an integral with a variable upper limit, F(x) = ∫ₐˣ f(t) dt. It records the signed area from a to x, and where f is continuous, F′(x) = f(x).
Net change theorem: The integral of a rate of change over an interval equals the net change in the quantity: ∫ₐᵇ F′(t) dt = F(b) − F(a). Decreases count as negative, so net change can differ from the total amount of change.
Antiderivative: A function whose derivative is the given function, such as x³ for 3x². Any two antiderivatives on one interval differ by a constant, which is why answers carry + C.
Definite integral: ∫ₐᵇ f(x) dx: the signed area between the graph of f and the x-axis from a to b, defined as the limit of Riemann sums. Area below the axis counts as negative.
Average value of a function: The integral of f over [a, b] divided by b − a: the height of the rectangle on [a, b] with the same signed area as f. A continuous function reaches its average value somewhere on the interval.

Work through an example

Let F(x) = ∫₀ˣ (t − 1) dt. Find a formula for F and its derivative, and explain where F decreases and why F(2) = 0.

Find an accumulation function and its slope →

Find a net change from a rate →

Find the average value of a function →

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